Problem 103 - Calculus 3

Question

Find the arc length of the curve on the given interval. \(\mathbf{r}(t)=t^{2} \mathbf{i}+\left(2 t^{2}+1\right) \mathbf{j}, \quad 1 \leq t \leq 3\)

Step-by-Step Solution

Answer

The arc length of the curve is \( 8\sqrt{5} \).

Step 1: Set up the Arc Length Formula

The formula for the arc length of a curve given by a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \) over an interval \( [a, b] \) is \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). For our vector function \( \mathbf{r}(t) = t^2 \mathbf{i} + (2t^2 + 1) \mathbf{j} \), we have \( x(t) = t^2 \) and \( y(t) = 2t^2 + 1 \).

Step 2: Find Derivatives

Calculate the derivatives: \( \frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t \) and \( \frac{dy}{dt} = \frac{d}{dt}(2t^2 + 1) = 4t \).

Step 3: Substitute into the Arc Length Formula

Substitute the derivatives into the arc length formula to get: \[ L = \int_{1}^{3} \sqrt{(2t)^2 + (4t)^2} \, dt = \int_{1}^{3} \sqrt{4t^2 + 16t^2} \, dt = \int_{1}^{3} \sqrt{20t^2} \, dt \].

Step 4: Simplify the Integral

Simplify the integral expression: \[ \int_{1}^{3} \sqrt{20t^2} \, dt = \int_{1}^{3} \sqrt{20} \cdot |t| \, dt \]. Since \( t \) is positive over the interval \( [1, 3] \), \( |t| = t \). The expression becomes \( \sqrt{20} \int_{1}^{3} t \, dt \).

Step 5: Evaluate the Integral

Evaluate the integral: \[ \sqrt{20} \int_{1}^{3} t \, dt = \sqrt{20} \left[ \frac{t^2}{2} \right]_{1}^{3} = \sqrt{20} \times \left( \frac{3^2}{2} - \frac{1^2}{2} \right) = \sqrt{20} \times \left( \frac{9}{2} - \frac{1}{2} \right) \].

Step 6: Calculate the Result

Compute the result: \[ \sqrt{20} \times \frac{8}{2} = \sqrt{20} \times 4 = 4\sqrt{20} \]. Simplify \( \sqrt{20} = 2\sqrt{5} \), so the arc length is \( 8\sqrt{5} \).

Key Concepts

Vector Calculus

In vector calculus, we often work with vector functions, which are functions that take real numbers and return vector quantities. These can represent paths or curves in two or three dimensions. For instance, a vector function like \(\mathbf{r}(t) = t^2 \mathbf{i} + (2t^2 + 1) \mathbf{j}\) describes the position of an object in a plane at a given time \(t\). Understanding vector calculus is crucial as it allows us to analyze physical phenomena that involve directions and magnitudes, such as velocity and force.

• It extends traditional calculus to handle vector input and output.
• It is foundational to topics like electromagnetism and fluid dynamics.

With the given vector function, we navigate the path by analyzing each component individually, providing an insight into how the whole system behaves.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function. In the context of our arc length problem, we need to find the derivatives of the component functions of the vector \(\mathbf{r}(t)\). This involves calculating \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), which tell us how the position changes with respect to time.

• The derivative \(\frac{dx}{dt} = 2t\) tells us how the \(x\)-component of the position changes as \(t\) changes.
• Similarly, \(\frac{dy}{dt} = 4t\) gives us the rate of change of the \(y\)-component.

These derivatives allow us to substitute back into the arc length formula. Mastering derivatives is essential as they are pivotal in optimization problems, curve sketching, and understanding motion.

Definite Integral

A definite integral provides the area under a curve, acting as a summation of infinite small quantities. In the arc length formula, we use a definite integral to "sum up" infinitely small linear segments of our curve to find the total length. Here, the formula \(L = \int_{1}^{3} \sqrt{(2t)^2 + (4t)^2} \, dt\) involves integrating with respect to \(t\), over the interval \([1, 3]\).

• We simplify the expression inside the square root to \(\sqrt{20t^2}\).
• This simplifies further to \(\sqrt{20} \int_{1}^{3} t \, dt\).

When evaluated, the integral reflects the total arc length of the curve. Definite integrals are key in many areas of mathematics and engineering, giving precise measurements of quantities like areas, volumes, and lengths.

Curves

Curves are smooth, continuous lines or paths on a graph, and they are often described using parametric equations or vector functions. When finding arc lengths, we are essentially measuring the distance along a curve between two points. Our given vector function \(\mathbf{r}(t)\) maps out such a curve in the Cartesian plane.

• Understanding curves involves analyzing their geometric properties like length, curvature, and tangents.
• In this case, we use the vector and its derivatives to precisely determine the curve's properties over a specified interval.

Analyzing curves is fundamental in fields like physics, engineering, and computer graphics, where precise modeling is necessary.